Sketching a phase portrait and proving it has no periodic orbits

Question

I'm doing this exercise:

Consider the following differential equation system:

$$\begin{cases}\dot{x}=x(1-x-ay) \\\dot{y}=y(1-y-bx)\end{cases}$$

With $a,b>0$ and different to $1$. Sketch the phase portrait on the first quadrant and prove it has no periodic orbits.

What I've done so far is to find the equilibrium points. We make

$$\begin{cases}0=x(1-x-ay) \\0=y(1-y-bx)\end{cases}$$

and we find equilibrium points at $(0,0), (1,0), (0,1)$ and at the intersection point between $x=1-ay\ $ and $y=1-bx$, wich is

$$x=\frac{1-a}{1-ab},\ y=\frac{1-b}{1-ab}.$$

Once we have this, we have to check what's the local behaviour og the orbits near every equilibrium point:

We first find $$J=\begin{pmatrix} -x+(-1+x+ay) & -xa\\ -yb & -y+(-1+y+bx) \end{pmatrix},$$

but checking every equilibrium point $(J_{(0,0)},J_{(1,0)},J_{(0,1)},\dots )$ doesn't give me nothing.

Am I missing something? Once I have the local behavior on the equilibrium points, how can I prove it has no periodic orbits?

Thanks for your time.

Answer 1

Do you have any conditions on the values of $a,b$? Calculating the Jacobian I got, $$ J\left( x,y \right)=\left[ \begin{matrix} 1-2x-ay & -ax \\ -by & 1-bx-2y \\ \end{matrix} \right] $$ Then $ J\left( 0,0 \right)={{I}_{2\times 2}} $ so the origin is unstable. Let $\sigma(A)$ denote the set of eigenvalues of $A$. Then,

$$ \sigma \left( J\left( 0,1 \right) \right)=\left\{ -1,1-a \right\}\quad \text{and}\quad \sigma \left( J\left( 1,0 \right) \right)=\left\{ -1,1-b \right\} $$

so for $a,b >1$ these equilibria will be stable. Finally,

$$\sigma \left( J\left( \frac{a-1}{ab-1},\frac{b-1}{ab-1} \right) \right)=\left\{ \frac{(a-1)(b-1)}{ab-1},\frac{1-ab}{ab-1} \right\} $$

If $a>1, b>1$ then the first eigenvalue here is positive; so this one is unstable.

Here is a plot for $a=b=2$ with the equilibria marked in red.

Answer 2

Calling

$$ x_0 = \frac{1-a}{a b - 1}\\ y_0 = \frac{1-b}{a b - 1} $$

We have the behavior near $(x_0,y_0)$

$$ \dot X = J_0 X $$

with

$$ X = (x-x_0,y-y_0)^{\top} \ \ \mbox{and}\ \ J_0 = \left(\begin{array}{cc} \frac{1-a}{ab-1} & \frac{a(1-a)}{ab-1}\\ \frac{b(b-1)}{ab-1} & \frac{1-b}{ab-1}\end{array}\right) $$

Analyzing the $J_0$ eigenvalues we obtain

$$ \{\frac{(a-1)(b-1)}{ab-1},-1\} $$

then $(x_0,y_0)$ never can be a center.

Also there are no periodic limit cycles because applying the Bendixson-Dulac criteria with $\phi(x,y) = x^{\alpha}y^{\beta}$

we have

$$ (\phi(x,y)x(1-x-a y))_x+(\phi(x,y)y(1-y-bx))_y = -\phi(x,y)((2+a+a\alpha+\beta)x+(2+\alpha+b+b\beta)y)-(2+\alpha+\beta)) $$

Now making

$$ \alpha = \frac{2-a(b+1)}{ab-1} $$

and

$$ \beta = \frac{2-b(a+1)}{ab-1} $$

we have

$$ (\phi(x,y)x(1-x-a y))_x+(\phi(x,y)y(1-y-bx))_y = \phi(x,y)\frac{2-a-b}{ab-1} $$

which is non null inside any of the four quadrants.

NOTE

The quadrant verification suffices because $x = 0$ and $y = 0$ are orbits.